Question 3

Where is the inverse of a point inside the circle?

The inverse of a point inside the circle is outside the circle:

the point lies on the ray (AR) going from the center of the circle (A) to the initial point (F)

:where A is the center of the circle and R is on the circle

Because F is inside the circle, AF is less than 1, but greater than zero

We know: (AF') = (AR)(AR) / (AF)

so (AF') = (1) / (a number less than 1, but greater than zero)

therefore, (AF') is greater than 1, which puts the point F' ( the inverse of F)

outside of the circle

Where is the inverse of a point outside the circle?

The inverse of a point outside the circle is inside the circle:

the point lies on the ray going from the center of the circle to the initial point

:where A is the center of the circle and R is on the circle

Becuase F is outside the circle, AF is greater than 1

We know: (AF') = (AR)(AR) / (AF)

so (AF') = (1) / (a number greater than 1)

therefore, (AF') is less than 1, which puts the point F' ( the inverse of F)

inside of the circle

Are there points that are their own inverses?

Yes, points that lie on the circle of inversion are there own inverses

Let AR (the radius of the circle of inversion) = 1

:where A is the center of the circle and R is on the circle

Because F is on the circle AF = AR = 1

We know: (AF') = (AR)(AR) / (AF)

so (AF') = (1) / (1) = 1

therefore, F' ( the inverse of F) must also lie on the circle of inversion

Where is the inverse of the center of the circle?

The inverse of the center of the circle is at infinity.

Let AR (the radius of the circle of inversion) = 1

: where A is the center of the circle and R is on the circle

Becuase F is on the center of the circle AF = 0

We know: (AF') = (AR)(AR) / (AF)

so (AF') = (1) / (0) = infinity

therefore, F' ( the inverse of F) is at infinity